Question: A parabola with equation $y=x^2+bx+c$ passes through the points $(-1,-11)$ and $(3,17)$. What is $c$?
Answer: We substitute these two points into the given equation to solve for $c$. Plugging in $(-1,-11)$, we get $-11=(-1)^2-1b+c\Rightarrow -b+c=-12$. Plugging in $(3,17)$, we get $17=3^2+3b+c \Rightarrow 3b+c=8$. In summary, we have the two equations \begin{align*}
-b+c&=-12\\
3b+c&=8
\end{align*} Multiplying the first equation by 3, we have $-3b+3c=-36$. Adding the second equation to this last one, we have $(-3b+3c)+(3b+c)=-36+8 \Rightarrow c=\boxed{-7}$.

The parabola is graphed below: [asy]

Label f;

f.p=fontsize(4);

xaxis(-9,4,Ticks(f, 2.0));

yaxis(-13,17,Ticks(f, 2.0));

real f(real x)

{

return x^2+5x-7;

}

draw(graph(f,-8,3), Arrows(4));

[/asy]